Question

$Evaluate the following\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \large\color{blue}lim_{n \to \: \infty }( \frac{1}{1 + \sqrt{1} } + \frac{1}{2 + \sqrt{2n} } + \frac{1}{3 + \sqrt{3n} } + - -\ \textless \ br /\ \textgreater \ \color{green}--- + \frac{1}{n + \sqrt{ {n}^{2} } }$
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Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{n \to \infty} \bigg[ \frac{1}{1 + \sqrt{1} } + \frac{2}{2 + \sqrt{2n} } + \frac{1}{3 + \sqrt{3n} } + . . . . . + \frac{1}{n + \sqrt{n^{2} } } \bigg]$

As we know that,

Using the concept of,

Definite integrals as a limit of sum.

We can write equation as,

$\sf \implies \displaystyle \lim_{n \to \infty} \sum_{r = 1}^{n}\bigg[ \frac{1}{r + \sqrt{rn} } \bigg]$

$\sf \implies \displaystyle \lim_{n \to \infty} \sum_{r = 1}^{n}\bigg[ \frac{1}{(r/n)n + \sqrt{(r/n)n^{2} } } \bigg]$

Replace : r/n = x, ∑ = ∫, 1/n = dx.

$\sf \implies \displaystyle \lim_{n \to \infty} \sum_{r = 1}^{n}\bigg[\frac{1}{(x)n + \sqrt{(x)n^{2} } } \bigg]$

$\sf \implies \displaystyle \lim_{n \to \infty} \sum_{r = 1}^{n}\bigg[\frac{1}{(x)n + n\sqrt{(x) } } \bigg]$

$\sf \implies \displaystyle \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n}\bigg[\frac{1}{(x) + \sqrt{(x) } } \bigg]$

We can write equation as,

$\sf \implies \displaystyle \int\limits^1_0 \frac{dx}{x + \sqrt{x} }$

As we know that,

By using substitution method in the equation, we get.

Let we assume that,

⇒ √x = t.

Squaring on both sides of the equation, we get.

⇒ x = t².

Differentiate w.r.t x, we get.

⇒ dx = 2tdt.

Now, we can write equation as,

$\sf \implies \displaystyle \int\limits^1_0 \frac{2tdt}{t^{2} + t }$

$\sf \implies \displaystyle 2\int\limits^1_0 \frac{tdt}{t(t + 1) }$

$\sf \implies \displaystyle 2\int\limits^1_0 \frac{dt}{(t + 1) }$

$\sf \implies \displaystyle 2 \bigg[ ln |t + 1| \bigg]_{0}^{1}$

Put the value of t = √x in the equation, we get.

$\sf \implies \displaystyle 2 \bigg[ ln |\sqrt{x} + 1| \bigg]_{0}^{1}$

As we know that,

In definite integration first we put upper limits then we put lower limits in the equation, we get.

$\sf \implies \displaystyle 2 \bigg[ ln |\sqrt{1} + 1| \bigg] - 2 \bigg[ ln|\sqrt{0} + 1 | \bigg]$

$\sf \implies \displaystyle 2 \bigg[ ln 2 \bigg] - 2 \bigg[ ln 1 \bigg]$

$\sf \implies \displaystyle 2(ln 2)$

$\sf \implies \displaystyle \boxed{ \lim_{n \to \infty} \bigg[ \frac{1}{1 + \sqrt{1} } + \frac{2}{2 + \sqrt{2n} } + \frac{1}{3 + \sqrt{3n} } + . . . . . + \frac{1}{n + \sqrt{n^{2} } } \bigg] = 2 ln(2)}$